Let $G$ be a group and $X$ set. $a,b \in X$ are in the same orbit, so show that $\mathrm{stab}(a) \cong \mathrm{stab}(b)$.

What I tried so far:

$a,b$ are in the same orbit. So that means $b \in \mathrm{orb}(a)$. That means there is $g'\in G$ so that $g'\cdot a=b$.

$\mathrm{stab}(a)=\{g \in G| g\cdot a=a \}$

$\mathrm{stab}(b)=\{g \in G|g\cdot b = b \}=\{g \in G|g\cdot g'\cdot a=b\}=\{g \in G|gg'\cdot a=b\}=\{g \in G|g.a=b\}$

So the second step $\{g \in G|gg'\cdot a=b\}$ is because of the properties of group action. The third step $\{g \in G|gg'.a=b\}=\{g \in G|g.a=b\}$ is because of the properties of group (so you can "get" to any element, with $g'$ being fixed element and $g$ is arbitrary). Are my steps so far correct? How should I continue?

Any hints and assistance would be much appriciated!


2 Answers 2


Let $ga=b$. Then $h \in stab(a)$ iff $ghg^{-1}\in stab(b)$, since $$ghg^{-1} b=gh a$$
now $$gha=b$$ iff $$ha=g^{-1}b=a$$

So $g stab(a)g^{-1}=stab(b)$.

  • $\begingroup$ Just to make sure, does saying "a,b are at the same orbit", mean $b \in orb(a)$? $\endgroup$
    – dsfsf
    Jun 23, 2014 at 23:41
  • $\begingroup$ Yes ga=b for some g $\endgroup$ Jun 23, 2014 at 23:45

Here is a very slick proof (in my opinion) of ${\rm Stab}(gx)=g{\rm Stab}(x)g^{-1}$.

Observe $(\color{Blue}{g{\rm Stab}(x)g^{-1}})(\color{Green}{gx})=g{\rm Stab}(x)x=\color{Green}{gx}\implies \color{Blue}{g{\rm Stab}(x)g^{-1}}\subseteq {\rm Stab}(\color{Green}{gx})$.

For the reverse containment, rearrange $\color{Purple}{g^{-1}}{\rm Stab}(\color{Green}{gx})\color{Purple}{g}\subseteq{\rm Stab}(\color{Purple}{g^{-1}}\color{Green}{gx})$.


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